Diffusion of water isotopes in polar firn¶
Image("../figs/dye3_highres_d18_profile.png", width = 1000)
Image("../figs/diffusion_bulb.png", width = 800)
Consider the two big bulbs of the figure above connected with a thin capillary. Both are kept in ssame pressure and temperature and have exactly the same volume. At $t=0$ one bulb contains only $\text{CO}_2$ and the other only contains $\text{N}_2$. For $t>0$ the gases start to mix. Looking into the evovling concentration of $\text{CO}_2$ in the right bulb we define the following:
- $\text{CO}_2 \text{ flux} = \left(\frac{\text{amount of gas removed}}{\text{time}}\right)$ $\left[\text{mol s}^{-1}\text{ m}^{-2}\right]$
In the Fickian approach the flux is dependant on the length of the capillary ie.
- $\text{CO}_2 \text{ flux} = D\left(\frac{\text{CO}_2\text{ concentration difference}}{\text{capillary length}}\right)$
The proportionality constant $D$ here is the diffusion coefficient
Diffusion across a thin film - Fick's 1st law¶
Image("../figs/ficks1stlaw.png", width = 1000)
Fick's 2nd law¶
Image("../figs/ficks2law.png", width = 1000)
Diffusion in polar firn¶
The porous medium of the firn allows for a diffusive process from te time of deposition until pore close-off. Mass transport along the isotopic gradients occurs attenuating the isotopic signals. Often this obliterates the annual cycles completely.
We can set up the diffusion equation taking into consideration thhe vertical compaction which in our case we will consider only to be due to firn densification (thus ice flow thinning is considered negligible) $\frac{\partial \delta}{\partial t} = D \left( t \right) \frac{\partial^2 \delta}{\partial z^2} - \dot{\varepsilon}_z \left( t \right) z ~\frac{\partial \delta}{\partial z} \enspace .$
The equation above can be solved with Fourier integrals. If the initial isotopic profile at $t = 0$ is $\delta \left(z, 0\right)$ then the solution for time $t'$ will be the convolution of $\delta \left(z, 0\right)$ with a Gaussian kernel $\mathcal{G} = \frac{1}{\sigma \sqrt{2\pi}} \, e^{\frac{-z^2}{2 \sigma^2}} \enspace$.
$\delta \left( z, t \right) = \mathcal{S} \left( t \right) \frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{+\infty} \delta \left( z, 0 \right) \exp \left\{ \frac{-\left(z-u \right)^2} {2 \sigma^2}\right\} \,\mathrm{d}u \enspace$.
Here $\mathcal{S}(t)$ is the total thinning equal to $\mathcal{S} \left( t' \right) = e^{\int_0^{t'} \dot{\varepsilon}_z \left( t \right) \mathrm{d}t} \enspace$ and $\sigma$ is the standard deviation of the Gaussian filter also referred to as the diffusion length.
A differential equation for the estimation of the diffusion length is:
$\frac{\mathrm{d}\sigma^2}{\mathrm{d}t} - 2\,\dot{\varepsilon}_z\!\left( t \right) \sigma^2 = 2D\!\left( t \right) \enspace$.
If we assume the simple strain rate $\dot{\varepsilon}_z \left( t \right) = -\frac{\mathrm{d\rho}}{\,\,\,\mathrm{d t}}\,\frac{\,1\,}{\,\rho\,}$ then a solution for the diffusion length can be given by $\sigma^2 \left( \rho \right) = \frac{\,1\,}{\rho^2}\int_{\rho_o}^{\rho}2\rho^2 {\left( \frac{\mathrm{d}\rho}{\mathrm{d}t}\right)}^{-1}\! D \!\left( \rho \right) \,\mathrm{d}\rho$
Diffusion will cause annuals to dissapear. This can be reversed as we will see.¶
Image("../figs/diffusion_withlaki.png", width = 1000)
Ice flow thinning - vertical strain rate¶
Image("../figs/thinning_strain.png", width = 1000)
Simple vertical strain rate¶
Image("../figs/simple_strain_rate.png", width = 1000)
One simple solution for diffusion lengths¶
Image("../figs/diffusion_length_solution_time.png", width = 1000)
A slightly more complicated solution for diffusion lengths¶
Image("../figs/diffusion_length_solution.png", width = 1000)
The firn diffusivity¶
Image("../figs/diffusivity_firn_holme.png", width = 900)
